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Chapter 3: Problem 100

A certain metal oxide has the formula MO, where \(\mathrm{M}\) denotes the metal. A 39.46-g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, \(31.70 \mathrm{~g}\) of the metal is left over. If \(\mathrm{O}\) has an atomic mass of 16.00 amu, calculate the atomic mass of \(\mathrm{M}\) and identify the element.

Short Answer

Expert verified
The atomic mass of metal 'M' is approximately 65.36 atomic mass units (amu). Referring to a periodic table, this atomic mass is closest to the atomic mass of the element zinc, hence metal 'M' is likely to be Zinc.

Step by step solution

01

Calculate the Mass of Oxygen

To find the atomic mass of the metal, first calculate the mass of oxygen in the oxide. Do this by subtracting the mass of the metal after heating from the mass of the metal oxide before heating. This mathematically translates to: Mass of Oxygen = 39.46 g (initial mass of compound) - 31.70 g (mass of remaining metal) = 7.76 g.
02

Determine the Molar Mass of Oxygen

The atomic mass of oxygen is given as 16.00 atomic mass units (amu). This also corresponds to the molar mass (16.00 g), because one mole of oxygen has a mass of 16.00 grams.
03

Calculate Moles of Oxygen in the Compound

Now calculate the number of moles of oxygen in the compound by dividing the mass of oxygen by its molar mass. Number of moles = 7.76 g (mass of oxygen) / 16.00 g/mol (molar mass of oxygen) = 0.485 moles.
04

Calculate the Atomic Mass of Metal

Since the formula of the compound is MO, it implies that there is one mole of metal M for every mole of Oxygen O. Thus, the same number of moles calculated for oxygen will be the number of moles for the metal. The atomic mass of metal 'M' can be obtained from its definition: it is the mass of one mole of the metal. Mathematically, use the formula: Atomic mass of 'M' = mass of 'M' / moles of 'M' = 31.70 g (mass of 'M')/ 0.485 mol (moles of 'M') = 65.36 amu.

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Most popular questions from this chapter

The depletion of ozone \(\left(\mathrm{O}_{3}\right)\) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is $$ \mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{O}_{2}+\mathrm{NO}_{2} $$ If \(0.740 \mathrm{~g}\) of \(\mathrm{O}_{3}\) reacts with \(0.670 \mathrm{~g}\) of \(\mathrm{NO}\), how many grams of \(\mathrm{NO}_{2}\) will be produced? Which compound is the limiting reagent? Calculate the number of moles of the excess reagent remaining at the end of the reaction.

On what law is stoichiometry based? Why is it essential to use balanced equations in solving stoichiometric problems?

How many moles of calcium (Ca) atoms are in \(77.4 \mathrm{~g}\) of \(\mathrm{Ca} ?\)

Define limiting reagent and excess reagent. What is the significance of the limiting reagent in predicting the amount of the product obtained in a reaction? Can there be a limiting reagent if only one reactant is present?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).
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